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Show your work: x - 8 = 12

5,0 is not a solution to the system of linear inequalities.

Answer: 3

Step-by-step explanation: The number in front of your variable is called your coefficient so we say that 3 is the coeficient.

Make sure to understand that a variable

is just a letter that represents any number.

Answer:

0.875?

Step-by-step explanation:

I just did 7 divided by 8

Answer:

14/16

49/56

Step-by-step explanation:

I looked it up for you!!!

Answer:

48

Step-by-step explanation:

Cat = x

Dog = 8x

x + 8x = 54

9x = 54

x = 6, Dog = 8 * 6 = 48

Answer:

Part A:

I would multiply the length, width, and depth of the pool to find the volume of it. That would be how much water can fit inside. 25 ft x 15 ft x 10 ft = 3750 ft

Part B:

a. 3750 ft

Answer:

f(x) = a(x - 1) + 20

Step-by-step explanation:

Vertex form is written in this format: f(x) = a(x - h) + k. The vertex point is written in the form (h,k). To write the vertex in the equation, fill in the digits where they belong.

Hope it helps!

The ball's parabolic route is represented by the quadratic function's graph. The quadratic function is x²-2x+4=0.

Given that,

A thrown ball's height is a quadratic function of the amount of time it has been in the air. The ball's parabolic route is represented by the quadratic function's graph. The ball's route includes the location (1, 20), which is the graph's vertex (0,4).

We have to find what does the expression for this function look like? In vertex form, write the quadratic function.

Use y= a(x-h)² + k as the vertex form of a parabola.

(The parabola opens upward if "a" is positive and downward if "a" is negative.) Additionally, the vertex of the (h,k) is a parabola.

The vertex of the parabola (h,k) is (1,20)

The a value is 4 and y as 0.

y=a(x-h)²+k

0=4(x-1)²+20

0=4(x²-2x+1)+20

0=4x²-8x-4+20

0=4x²-8x+16

4x²-8x+16=0

x²-2x+4=0

Therefore, the quadratic function is x²-2x+4=0.

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Answer:

Ur answer is in attachment

Step-by-step explanation:

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The probability that a student has both box and a console is 1/6 and the P(B/A) = 0.378

Probability is a measure of the likelihood of an event occurring. The probability formula is defined as the possibility of an event happening being equal to the ratio of the number of favorable outcomes and the total number of outcomes. Probability of event to happen P(E) = Number of favorable outcomes/Total Number of outcomes.

Given here, the number of students with both video games is 30 thus

The probability that a student has both box and a console is 30/180=1/6

And the conditional probability P(B/A) is = 1/6/0.44

                                                                    = 0.378

Hence, The probability that a student has both box and a console is 1/6, and the P(B/A) = 0.378

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Answer:

21

Step-by-step explanation:

Answer:

x = 16

Step-by-step explanation:

[tex] \frac{2x - 5}{x + 8} = \frac{22.5}{20} [/tex]

Cross multiply

[tex] (2x - 5)*20 = (x + 8)*22.5 [/tex]

[tex] 40x - 100 = 22.5x + 180 [/tex]

Collect like terms

40x - 22.5x = 100 + 180

17.5x = 280

Divide both sides by 17.5

x = 280/17.5

x = 16

Answer:

Hey mate.....

Step-by-step explanation:

This is ur answer....

Hope it helps!

Mark me brainliest pls....

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Answer:

168 ways

Step-by-step explanation:

Given

[tex]Junior = 7[/tex]

[tex]Senior = 3[/tex]

Required

Number of ways 3 project can be assigned

Starting with the second project.

This can be assigned to only senior coders

So:

[tex]n_1 = 3[/tex]

Then, the third project.

This can be assigned to only junior coders

So:

[tex]n_2 = 7[/tex]

At this point, we are left with 2 senior coders and 6 junior coders

Since the first can be assigned to anybody.

[tex]n_3 =2+ 6[/tex]

[tex]n_3 =8[/tex]

The number of selection is then calculated as:

[tex]n = n_1 * n_2 * n_3[/tex]

[tex]n = 3 * 7 * 8[/tex]

[tex]n = 168[/tex]

Answer:

The appropriate solution is "1.7109". A further explanation is provided below.

Step-by-step explanation:

The given values are:

Sample size,

n = 25

Degree of freedom,

[tex]df=n-1[/tex]

   [tex]=25-1[/tex]

   [tex]=24[/tex]

At 90% confidence level,

[tex]\alpha=1-90 \ percent[/tex]

  [tex]=1-0.9[/tex]

  [tex]=0.1[/tex]

and,

[tex]\frac{\alpha}{2}=\frac{0.1}{2}[/tex]

  [tex]=0.05[/tex]

Now,

The correct values of t will be:

=  [tex]t_{\frac{\alpha}{2}, df }[/tex]

=  [tex]t_{0.05,24}[/tex]

=  [tex]1.7109[/tex]

Answer:

Step-by-step explanation:

i think that the equations might be y = x - 7

Answer:

26 square feet

Step-by-step explanation:

I dont know if this is right

Answer:

4

Step-by-step explanation:

Answer:

1/10 = 4  

Step-by-step explanation:

Answer:

d≈21.99

Step-by-step explanation:

Answer: W is your answer :)

Step-by-step explanation:

Answer:

m = 29

Step-by-step explanation:

You are counting by fives.

25, 30, 35

m + 1 = 30

m + 1 - 1 = 30 - 1

m = 29

25, (29 + 1), 35

25, 30, 35

The value of m is equal to 29

An arithmetic sequence is sequence of integers with its adjacent terms differing with one common difference.

An arithmetic sequence, therefore, is defined by two parameters, viz. the starting term and the common difference.

We are counting by fives.

m + 1 = 30

Now subtract 1 on both sides;

m + 1 - 1 = 30 - 1

m = 29

Thus substituting the value to find other terms;

25, (29 + 1), 35

The sequence is

25, 30, 35

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Answer:

-2, 1, 4

Step-by-step explanation:

Answer:

Equation: [tex]143 + g = 180[/tex]

[tex]g =37[/tex]

Step-by-step explanation:

Given

The attached figure

Required

The equation to find g and the solution

From the attached: Angles ACD, DCE and ECB lie on a straight line.

Where

[tex]ACD = 90[/tex]       [tex]DCE = 53[/tex]     [tex]ECB = g[/tex]

This means that:

[tex]ACD + DCE + ECB = 180[/tex]

[tex]90 + 53 + g = 180[/tex]

[tex]143 + g = 180[/tex]

Subtract 143 from both sides

[tex]-143 + 143 + g = 180 - 143[/tex]

[tex]g = 37[/tex]

Answer:

b

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]T_j \sim exp ( \lambda j) \ \ \ j = 1 (1) n \\ \\ f_{Tj}(tj) = \lambda j e^{-\lambda j tj } , tj>0 \\ \\ P\Big[ Tj> x\Big] = \int \limits ^{\infty}_{x} \lambda j e^{-\lambda j tj}\ dtj \\ \\ = e^{-\lambda j x}, x > 0 \\ \\ \\ \\ a) F_x (x) \\ \\ = P[X \le x ] \\ \\ = 1 - P[X> x] \\ \\ = 1 - \pi \limits ^{n}_{j =1} \Big\{ P[T_j > x ] \Big \}[/tex]

[tex]\text{This is because the appliance has the capacity to work for more than (x) }[/tex][tex]\text{hours if and only if all the "n" components work more than (x) hours.}[/tex]

[tex]\text{Then:}[/tex]

[tex]= 1 - e \ \pi^{n}_{j=1} \ e^{-\lambda j x} \\ \\ = 1 - e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}[/tex]

∴

[tex]CDF = 1 - e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x>0[/tex]

[tex]PDF =\Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x>0[/tex]

[tex]f_x(x) = \Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x>0[/tex]

(b)

[tex]E(x) = \int \limits ^{\infty}_{o }x f_x (x) \ dx \\ \\ = \int \limits ^{\infty}_{o }x \ \Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ dx[/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j=1} \lambda j} \ \ \int \limits ^{\infty}_{o} \Big [(\sum \limits ^n_{j=1} \lambda j )x \Big] e ^{-\Big ( \sum \limits ^{n}_{j=1} \lambda j \Big)x} \ \ d \Big( x \sum \limits ^n_{j=1} \lambda j \Big)[/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j} \int \llimits ^{\infty}_{o} t e^{-t} \ dt[/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j} \int \llimits ^{\infty}_{o} t e^{-t} \ dt \ \ \ \ \text{\Big[By transformation }t =( \sum \limits ^n_{j=1} \lambda j )x \Big][/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j}[/tex]

(c)

[tex]Let \ \ l_y_{\dfrac{1}{2}} \text{ be the median}[/tex]

∴

[tex]F(l_y_{\dfrac{1}{2}}) = \dfrac{1}{2} \\ \\ 1 - e ^{-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big)} {l_y}_{\frac{1}{2}} = \dfrac{1}{2} \\ \\ \dfrac{1}{2} = e^{-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big) } {l_y}_{\frac{1}{2}} \\ \\ - In 2 = {-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big)} {l_y}_{\frac{1}{2}} \\ \\ \\ \\ \mathbf{ {l_y}_{\frac{1}{2}} = \dfrac{ In \ 2}{\sum \limits ^{n}_{j=1} \lambda j }}[/tex]

Answer:

sphere

Step-by-step explanation:

sphere has a larger volume

Answer:

C, 33 1/3%

Step-by-step explanation:

Because there are only two even number that follow this rule: 2<x≥6, and since there are only 6 possible outcomes, the probabilty is 2/6, which is 1/3. In a percent form, this is 100%*2/3, or 33 1/3%.

Answer:

98%

Step-by-step explanation:

Answer:

2/12 or 1/6

Step-by-step explanation:

12 games total

they have 2 games that are there favorite

Answer:

16

Step-by-step explanation:

8x2